php - SQL 查询的语法错误

Question

下面是代码，但我收到此错误

You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near '' at line 3

任何帮助将不胜感激。

 <?php
 include("../includes/conn.php");
 $sql = "SELECT * 
 FROM  `t_maincontent`
 where uid_page =". $_GET['page'] ;
 $results = mysql_query($sql, $conn) or die(mysql_error());
 $content = mysql_fetch_assoc($results);
 //print $content['nv_content']."¬".$content['nv_title']."¬".$content['nv_meta'];
 ?>

Answer 1

请阅读有关从 PHP 进行数据库查询的基础知识。

您应该检查是否 $_GET['page']通过 isset( $_GET['page'] ); 存在仅发送查询(如果存在)。

include( '../includes/conn.php' );
if( isset( $_GET['page'] ) )
{
    /* cast to int would be enough to cancel SQL injection attacks */
    $sql = 'SELECT * FROM  t_maincontent WHERE uid_page = ' . (int) $_GET['page'] ;
    $results = mysql_query($sql, $conn) or die(mysql_error());
    $content = mysql_fetch_assoc($results);
}